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<blockquote data-quote="mylove1997" data-source="post: 156948" data-attributes="member: 270107">vì M(2;1) nên gọi dường thẳng cần tìm có dạng:<img src="https://codecogs.izyba.com/png.latex?\150dpi%20(\Delta%20):a(x-2)+b(y-1)=0%20;a^{2}+b^{2}\neq%200" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?\150dpi%20\Rightarrow%20ax+by-2a-b=0" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />Ta có <img src="https://codecogs.izyba.com/png.latex?\150dpi%20(\Delta):%20VTPT%20n_{1}(a;b);%20(d):VTPT%20n_{2}(2;3)" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?\150dpi%20cos(\Delta%20,d)=\left|cos(n_{1};n_{2}%20)\right|=\frac{\left|n_{1}%20.n_{2}\right|}{\left|n_{1}%20\right|.\left|n_{2}%20\right|}=\frac{\left|2a+3b%20\right|}{\sqrt{13(a^{2}+b^{2}}}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />Mà<img src="https://codecogs.izyba.com/png.latex?\150dpi%20(\Delta%20,d)=45^{0}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?\150dpi%20\Rightarrow%20\frac{\left|2a+3b%20\right|}{\sqrt{13(a^{2}+b^{2}}}=\frac{\sqrt{2}}{2}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?\150dpi%20\Rightarrow%20-10a^{2}+10b^{2}+48ab=0" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />Chọn b=1 suy ra a=-1/5 hoặc a=5Vậy<img src="https://codecogs.izyba.com/png.latex?\150dpi%20(\Delta%20):\frac{-1}{5}x%20+y-\frac{3}{5}=0" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /> hay <img src="https://codecogs.izyba.com/png.latex?\150dpi%20-x+5y=3=0" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /> hoặc<img src="https://codecogs.izyba.com/png.latex?\150dpi%20(\Delta%20):5x+y-11=0" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /></blockquote>

[QUOTE="mylove1997, post: 156948, member: 270107"] vì M(2;1) nên gọi dường thẳng cần tìm có dạng:[IMG]https://codecogs.izyba.com/png.latex?\150dpi%20(\Delta%20):a(x-2)+b(y-1)=0%20;a^{2}+b^{2}\neq%200[/IMG] [IMG]https://codecogs.izyba.com/png.latex?\150dpi%20\Rightarrow%20ax+by-2a-b=0[/IMG] Ta có [IMG]https://codecogs.izyba.com/png.latex?\150dpi%20(\Delta):%20VTPT%20n_{1}(a;b);%20(d):VTPT%20n_{2}(2;3)[/IMG] [IMG]https://codecogs.izyba.com/png.latex?\150dpi%20cos(\Delta%20,d)=\left|cos(n_{1};n_{2}%20)\right|=\frac{\left|n_{1}%20.n_{2}\right|}{\left|n_{1}%20\right|.\left|n_{2}%20\right|}=\frac{\left|2a+3b%20\right|}{\sqrt{13(a^{2}+b^{2}}}[/IMG] Mà[IMG]https://codecogs.izyba.com/png.latex?\150dpi%20(\Delta%20,d)=45^{0}[/IMG] [IMG]https://codecogs.izyba.com/png.latex?\150dpi%20\Rightarrow%20\frac{\left|2a+3b%20\right|}{\sqrt{13(a^{2}+b^{2}}}=\frac{\sqrt{2}}{2}[/IMG] [IMG]https://codecogs.izyba.com/png.latex?\150dpi%20\Rightarrow%20-10a^{2}+10b^{2}+48ab=0[/IMG] Chọn b=1 suy ra a=-1/5 hoặc a=5 Vậy[IMG]https://codecogs.izyba.com/png.latex?\150dpi%20(\Delta%20):\frac{-1}{5}x%20+y-\frac{3}{5}=0[/IMG] hay [IMG]https://codecogs.izyba.com/png.latex?\150dpi%20-x+5y=3=0[/IMG] hoặc[IMG]https://codecogs.izyba.com/png.latex?\150dpi%20(\Delta%20):5x+y-11=0[/IMG] [/QUOTE]

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